HW#5%20solution%20corrected - V 15 m/s.— Area = 0.3 m2...

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Unformatted text preview: V: 15 m/s .— Area = 0.3 m2 r \ ,.. \‘ Area=1m2 CV .42" if}; Water flows as two free jets from the tee attached to the pipe shown in Fig. P533. The exit speed is 15 m/s. If viscous effects and gravity are negligible, determine the x and M” y components of the force that the pipe exerts on the tee. . _ , Area z 0.5 m2 M56 ‘H’Ié COHfi/a/ volume, Shown. Far 'H’Ie X—cam’aonem‘ of #6: 199m: fixer/{0’ by +116 Pllpé’. 0n 'f/ve +86. we use, #18. x— wmpanenf of #26 Izhear manden7‘um egaafimn. “MFVAI4’VW0K’A3 :KA,'gA3—%5{Ad’qz)+f ‘féap‘zwflr/g; 42.,» MM) + a =..P 4. + I; (1') 9‘17: , n 76* ‘7 W use, conservav‘fan 07‘- mass 6?, 3 Q2 ‘I’ 623 M A,v,=fizvz +4314 _ 50 V’ = AZVL+A3V3 = W=/Zflflé A, /,,,1. 70 esffma/e. R9qie we. Hie Berna/fig 7%, {/W ée/Wm” (dam/(z) 4L. Baas? 4 VI = flange 4— V:— /a 3: 1 lo [sag 2; /z’.fl1](/ N. 5") K§Qx=/{’”2;"")=(””Z§-i g ‘21“) : 40,50011£ time In Now mm: 53.0) we 76’“ t — {2 9(7qu 45?sz 4. (5g)(?79§)fi5§%.5m2](/%:) = m? (40,500 I: Wine) + Fx l»?- CTY‘ ~72 WIN/2. F I K 50 a -.: 72/0WN é— F’h’m} (mm/Waf/fi; Arc-L emkdgy/hfl’kahmktWCuja/fv y (WmonyL ml fh //nea>r mmw/wm ermég, ,4, $.,/ I VlFl/z I42 :1 5 {KEJfiWEEJflSgJflzmj 2 67,400N 7‘ = F} 5.37 Water is sprayed radially outward over 180° as indicated + in Fig. P537. The jet sheet is in the horizontal plane. If the jet See‘h'onfi ) velocity at the nozzle exit is 20 ft/s, determine the direction and m, Wilma magnitude of the resultant horizontal anchoring force required ' A __ w to hold the nozzle in place i ‘t . FIGURE P5. 37 The confin/ va/umé Ihc [ad'es fl»: may]: and Wa/er bc/wcer; 58675an 0) am! {2) a; xhdicaed #3 7%,; skehaA aéave. AW/lufim of 7% y d/iecfiori campanenf 0W“ #1:. “may momenfimv eguafi'on yib/J5 77' z I or 1;” sv/Of(_ié6059)(vz)/1Rd9 = flhfigfimmsmo) o and F g 0 4,}: :2: Ala/7501190,; of fine X d/Fecfi'on Campanern’ of 1%: Una)» momenfixm eg act/ion lend: 7‘0 {5 (LIOV. 174/? s at X 77- 1 F = fl/(tgszwflvgkmg —.— flaw; (MSG—60:77) 19x ’ a ‘ 2 . I . fi' “"01 F _ (qu/u]5)(0v5in.) (81".) (205)/Z)/ /6 ) Ax , #3 (127$) (/23) 5/9.; OP AX ~--—— , Campus—fl 5.410 A variable mesh screen produces a lin- ear and axisymmetric velocity profile as indicated in Fig. P5.4.0 in the air flow through a 2~ft-- diameter circular cross section duct. The static: pressures upstream and downstream of the screen are 0.2 and 0.15 psi and are uniformly distributed over the flow cross section area. Neglecting the force exerted by the duct wall on the flowing air, "I = 0‘2 “5i "2 = 0'15 05‘ caicuiate the screen drag force. VI ‘ ‘00 "'5 FIGURE P540 i - -— L Will/IIIIIIIII {Willi/[11117]. Section (1) Section (2) APP/("Cain's of five axia/ “Myaneflf 0/ file I’Ihea/ mama/um eguafi'm 7‘0 five f/ow #troujh five Lon/ml Va/u/ne shown 1;; fire ska r‘o/«i lam/5 7‘0 R --l{(ol{;4[ +ft12€u2 Zfi‘rdr : EA ,_ Kx, PA 0 o . ' 7736 value of an,“ may be Oé'lfllfled {Mm comer-mill”: of may: as {ct/laws ' R 2 M 1:2 =p/(um 27mg” ’7‘ I R 0 77m: 5, = i4 D,“ 12 max 7'“— (Zjigy l’zdr O me / 1 2 2 7. = 40023,? flu moi-f (29} / AS -2170.Wf/_7 /§4f9{2fi L K" ( flair ( 1432/ 5 51 2 _ z _ + 0.2 I}; 7? (2H) 144117").fi/515)1r(28)144£) ‘ 4 H‘ = 3 VI =7 E l00£7=l507tf 2 . 5.453 In a laminar pipe flow that is fully det. average velocity, E, with the axial direction mo— veloped, the axial velocity profile is parabolic, thatis, r2 2 c 1.— —— u “l (M as is illustrated in Fig. P543. Compare the axial direction momentum flowrate calculated with the velocity distribution taken into account. FIGURE P543 7716 axia/ dlfecfion meme/714w” [/owmfc based on a um'émq magi}, pmfi/e wifh 6: = 12 [5 MI: :1 [l J :: “Z Z X’W’firm flu/4 flu ’lrl? 7718 axial direcy‘l‘on mower/firm [/owmfe based on Me non. uniform paraéa/io vc And/y pro fi/c x3 K / 7. MP :[u “Zwrdr : Mlzfl‘kz/Z, r-7/r r X170”- (0 fl 4; /—~ ~jd~ a 0 (a R 4) 7— 2 u 7/")? Min“? " W C (071/010?) 3 To own)» a Elwyn/1;); befween a 4440’ ug we use 7% conservm‘ion of mass egaafi'on as 7(0//aw.$ was/Mym + 0 77w; a z: 91:: Z M}: = 4 [7511’]? g f. M]: . {nm— 37 3 X’um m anhém. 5—43 mentum flowrate calculated with the nonuniform 5.53 The exit plane of a 0.20-m-diameter pipe is partially blocked by a plate with a hfllfi in it that produces a OJO-m—di- meter strum as shown in Fig. P553. The water velocity in the pipe is 5 m/s. Gravity and viscous effects are negligible. Dcwr- mine the force needed to hold the plate against the pipe. 5 mls IFIGUIRE P5.53 The x*compoflem‘ of #he Mame/Wm eql/ah'on For H76 oom'ra/ Volume show” is V, ‘ __ _ .E 9., _I F .3- I_. y V3. [uevhdmrx Mfg Lx E3 cs I._.. __ __ ___'(13 P150 v, p(-V.)fl. napvlnl m4 — F) "3 where F13 {/78 fine in bold {hep/01‘s. 7771/5J F = pfl, WWW), -(>Vz"flz =,o,fl, MHZ—V1) (n where 175 = e/ilV, =97? %[g(0:2m31] (5%) z 57% 14150J A,V, =Azlé 0,- V; = (A,/Az.)V, ‘49, /Dz)"\4=(o.2m/0./m)1(5%)=20m,<§ In add/5m, from H73 Bernoulli eqz/ah'm} 70, +-z'—€V,1 = flz+%€\/22 $0 Md MW] 70;“? Thu-3 from EMU F s [.87x105%[.g(0.2m)1] +157 é?- ( 5% “20%) = 3J5an 5.54 Two water jets of equal size and speed strike ewh other as shown in Fig. P554. Determine the speed, V, and direction, 6, 7 of the resulting combined jet. Gravity is negligible. x I FIGURE P5.54 For Hm Confrbl vokume Shown TM 4hr. Skd’o‘? above “Hm: Urea! moonth equafiOA For 11m: x and ydtmc-Hahs are, 'Foy' +he, x dt'recfiom szszAz +(VCOSGDQVA = O (I) (Md 1"Dv Hat )1 citric.th ~Vle‘A,+(v sinthA : 0 (2) Also “For conseva on 0? M55 we. have PIVIAI “f PM: " PM = 0 (3) From Ecbs. I and 2. we. qe+ Viki = (£0.56 = (0+9 WA. SM 6 1 $0 -| '1 _, z (0.1” 6 = «3“ V1“: = (30+, (WC-$37" T , = 45° WA: mfi)‘ W (w? L} NOW, wmb‘ning E%S.1 and 3 we gel- -Vz‘A‘a— Vsine (V,A,+\12AL) -.- O of '2. v: M. sin e (V, A, + 01 AL) V 3 min," (33H) _.-——— _.._—-——--— ‘1. (Sinqs‘, ) [00 fit) ’W(O.tfi)1’+<10 Tr (OJ-H) ] and ‘f ’+ V : 107 {i :3" 5 54:5 )5. 6% 5.65 A 3-insd-iameter horizontal jet of water stokes a flat plate as inclicated in Fig. P165. Determine the jet velocity if a l()--lb 3 in, ' horizontal force is required to (a) hold the plate stationary, ' F _ 10 lb (b) allow the plate to move at a constant speed of 10 ft/s to the A _ rig-ht. (a) (b) FIGURE 95.65 The comlro/ Volume Sham/n Ih f/ze 565ch I}: (Med. 721$ sfdfi'anary p/a/o can i: cam/Hereo/ fifn‘. A/y/z'cafi’on of Me hay/yonfa/ (W X« dlfecfibn Com/manenf' of file 0364,, momel/qu/VM egmflm/r y/e/dj "uh/our A] z ’51,): 0r é { M, =(/ 51") 3 FAX I” ’4/ r7 7:42 Thus if L [1 :7. M.“ k V z I __...___..__ {.94 {131: 17,8142: / ’5 - {1‘3 6‘(/2 1}). z 5/” ’9‘ a of /0 2 70 "7 j “‘2 V! (4 = o ' , E. 57(dfi0nay lo/ach When Me y/az‘e move; 1‘0 Me r/y/n‘ wifh a speed 1 U= /0 151‘ , ' .17 7%e xa direczé'on comanenf of 71/18 0): ear momem‘um 67414291”, y/e/dj —(uI—U),o(u,—U)/ll = ~63 0/” F 1;: IX L ‘uI—U=_L)=_F;LX) A 2 and IF log. 1 ‘ f #70, a =/___¢__'4X2 1» 'U’ _ f- {7‘ __ {2" I vote) - #12? 7L- /0? - 20.23. ’7‘ mall/h; p/m‘c ...
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This note was uploaded on 05/04/2010 for the course TAM 335 taught by Professor Whoever during the Spring '08 term at University of Illinois at Urbana–Champaign.

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HW#5%20solution%20corrected - V 15 m/s.— Area = 0.3 m2...

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